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n200080 [17]
2 years ago
6

The mass percent of oxygen in pure glucose, c6h12o6 is 53.3 percent. a chemist analyzes a sample of glucose that contains impuri

ties and determines that the mass percent of oxygen is 49.7 percent. which of the follow impurities could account for the low mass percent of oxygen in the sample?
Chemistry
1 answer:
Dafna1 [17]2 years ago
7 0
We can calculate the mass percent of an element by dividing its atomic mass by the mass of the compound and then multiply by 100:
     % by mass of element = (mass of element/mass of compound) x100%
Impurities like n-eicosane with the molecular formula C20H42 could account for the low percent by mass of oxygen in the sample because it has a zero percent oxygen based on its compound formula which indicates that it does not have the element oxygen.
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kumpel [21]

Answer:

D

Explanation:

solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M NaOH Can resist pH change when there is little addition of either acid or base, hence it is a buffer solution

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2 years ago
If Co(NH3)63+ has a λmax at 440 nm, calculate ΔE for the complex. A) 2.72 x 10-4 kJ/mol B) 4.52 x 10-2 kJ/mol C) 2.72 x 10 2 kJ/
riadik2000 [5.3K]

<u>Answer:</u> The energy of the complex is 2.72\times 10^2kJ

<u>Explanation:</u>

To calculate the energy of the complex, we use the equation given by Planck which is:

\Delta E=\frac{N_Ahc}{\lambda}

where,

\lambda = Wavelength of the complex = 440nm=4.40\times 10^{-7}m    (Conversion factor:  1m=10^9nm )

h = Planck's constant = 6.624\times 10^{-34}Js

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N_A = Avogadro's number = 6.022\times 10^{23}

\Delta E = energy of the complex

Putting values in above equation, we get:

\Delta E=\frac{6.022\times 10^{23}\times 6.624\times 10^{-34}\times 3\times 10^8}{4.40\times 10^{-7}}\\\\\Delta E=2.72\times 10^{5}J=2.72\times 10^2kJ

Conversion factor used:  1 kJ = 1000 J

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5 0
2 years ago
A gas occupies 22.4 l at stp and 14.5 l at 100c and 2.00 atm pressure. how many moles of gas did the system gain or lose?
azamat
<span>At standard temperature and pressure 22.4 l of an ideal gas would contain 1 mole. in order to find the change in moles we must look at the ideal gas law PV=nRT where P=Pressure V=volume n=Moles R= Gas constant T= Temperature. To simplify this equation we will be using the gas constant at .08206 L-atm/mol-K. We must first convert 100c to k which is 373.15. Then we can plug the values into our equation which gives us (2atm)(14.5 l)=(n)(.08206 L-atm/mol-K)(373.15). After some basic algebra we get the moles to equal roughly .95 which is .05 moles less than our original system.</span>
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AnnZ [28]

Answer : The correct options are,

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